Trignometric Ratios, Functions and Equations Test - 3

tan(660^o ) cot(1200^o )
⇒tan(720 - 60^o ) cot(1080+120^o )
⇒- tan60^o cot120^o
⇒- tan60^o (-cot60^o )
⇒1

As we know that : sin(180o −x) = sin(x)
Plug in x = 30 to get
sin(180o −30 °) = sin(30 °)
 ⇒1/2
sin(90+θ) = cos θ
sin (90+60)=cos60=1/2

Since, cos(−x) = cosx
⇒cos(−435 °) = cos(435 °)
= cos(360 °+ 75 °)
= cos(75o)
= cos(90 °−15 °)
= sin15 °

To solve this problem, we 'll use the tangent addition formula:

• tan(A + B) = (tanA + tanB) / (1 - tanA * tanB)

We can rewrite 45 °as 22 °+ 23 °. So,

• tan(45 °) = tan(22 °+ 23 °)

Using the tangent addition formula:

• 1 = (tan22 °+ tan23 °) / (1 - tan22 °* tan23 °)

Cross-multiplying, we get:

• 1 - tan22 °* tan23 °= tan22 °+ tan23 °

Rearranging the terms, we get:

• tan22 °+ tan23 °+ tan22 °.tan23 °= 1

Therefore, the value of tan22 °+ tan23 °+ tan22 °.tan23 °is 1.

sec 390 °
= sec(360 °+ 30 °)
= sec 30 °= 2/(√3)

cos(7 π/12)
= cos(3 π/12 + 4 π/12)
= cos(π/4 + π/3)
Use trig identity: cos (a + b) = cos a.cos b - sin a.sin b
cosa = cos(π/4) = √2/2;
cosb = cos(π/3) = 1/2
sina = sin(π/4) = √2/2;
sina = sin(π/3) = √3/2
cos(7 π/12) = cos(π/4 + π/3)
=>(√2/2)(1/2) −(√2/2)(√3/2) 
= (√2 −√6)/4
= √2(1 - √3)/2 √2 √2
= (1 - √3)/2 √2

tan210 °
= tan (180 °+ 30 °)
= tan30 °
= 1/√3

(cos135 °- cos120o)/ (cos135 °+ cos1200)
= cos(180 °- 45 °) - cos(180 °- 60 °)/cos(180 °- 45 °) + cos(180 °-60 °)
=  (-cos45 °+ cos60 °)/(- cos45 °- cos60 °)
=  (-1/√2 + 1/2)/(- 1 √2 - 1/2)
=  -2 + √2 / -2 - √2  
= (-2 + √2 / -2 - √2) * (-2 + √2)/(-2 + √2)
=  (-2 + √2)2/(-2)2 - (√2)2
= (4 + 2  - 4 √2) / 2
= (6 - 4 √2)/2
= 3 - 2 √2

Correct Answer : a

Explanation :  {1 + cot α- sec(π/2 + α)} {1 + cot α+ sec(π/2 + α)}

As we know that (a-b)(a+b) = a² - b²

(1 + cot α)² - [sec(π/2 + α)]²

1 + 2cot α+ cot² α- (-cosec α)²

2cot α+ 1 + cot² α- cosec² α

As we know that 1 + cot² α= cosec² α

= 2cot α+ cosec² α- cosec² α

= 2cot α